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Greco
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The flywheel of a motor is connected to the flywheel of an electric generator by a drive belt. The flywheels are of equal size
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Black hole in X-Ray Binary.
An X-ray binary consists of 2 stars with masses (the accreting compact object) and (the donor). The
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Answers (71)
a) 0.256
Goodnight and thank you all for your help !!!
P(final) = P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))
fred h=10km=10000m and temp.T= 273+20=293, M=0.029 alpha=0.0065 and M=0.029 and the formula for a) is P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))
thx a lot vivipop!!!
and vivipop rescues the day (or better night!)
it is midnight here in Greece and i'm feeling realy tired ....last call for help for b and c as i have only one try left..
fred to the same system but be careful with the units, give me temperatutr and i ll tell you a)
give your values and i ll tell you the answer
i need b and c for the problem above (20oC etc) as it is i have only one left i found a ) p=0.263
some help....anybody????
|ôP|= 1.745 |á|= 1.577 T= 1.477
.885 w^2 = 15 w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period
I = m L^2 + (2/3) m R^2 -(k L^2 + mgL)T = .885 d^2T/dt^2 well calculate the coefficient of T on the left e.g. -(7 + 8) T = .885 d^2T/dt^2 .885 d^2T/dt^2 = - 15 T Torque = -15T where T = 5 * pi/180 alpha = d^2T/dt^2 = - 15 T /.885
b and c
but what is the answer for b and in this problem a)po=0.263
P(final) =P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha)) green tick
x1= 1.58 t1= 0.392
|ôP|= 1.308 |á|= 1.478 T= 1.526
|ôP|=1.308 |á|= 1.478 T= 1.526
"ì(x)=áx, with á= 1.0 m−1" so alpha is 1.0 m^-1 i got a green tick so in this problem alpha is 1.0 m^-1. In mine was m g k R a 1 10 8 2 0.8 and the answer was x1=1.58 t1=0.59
how did you find alpha?
|ôP|= 1.83 |á|= 1.84 T= 1.37
Hawk, c) The block will move back and reach a second stop somewhere between x=0 and x=x1. is WRONG
(v_0^2*(1-sin(alpha))/(g*cos^2(alpha))
i got the green tick so here is the solution equation (1) m1gR=0,5m1v1^2+0,5m2v2^2 m2v2-m1v1=0 so equation (2) v2=m1v1/m2 we substitute to equation (1) m1gR=0,5m1v1^2+0,5m2(m1v1/m2)^2 and we solve for v1 Ps i expect a thanks ( xaxaxaxx)
a) 45+a/2=51 degrees b)(1−sin)*v0^2/(g*cos^2(á))
"d=sqrt(m Vi^2/(mg*alpha+k))" whats that alpha in your solution? shouldn't it be d=sqrt(m Vi^2/(mg*+k))?
and if the temperature was 20∘C?
i got the green tick so the right answer is a) 45 + (alpha/2) (45 is the max angle without incline plus the a/2 for the incline b) (v_0^2*(1-sin(alpha))/(g*cos^2(alpha))
The angle should be less than 45 degrees how is that possible if 45+a/2 is the correct answer then the angle is more than 45 degrees..
23220
a) I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19
v=113.14 a=0.90
a)25 b)2,55 c)9,27
I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19
I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19
I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19
i check them they are ok thx a lot..
I=1/3*m*L^2 Eini= mg(L/2) + 0 Efin= mg(L/2)cos30 + 1/2*I*w^2 Eini=Efin -> w=sqrt(3*g (1-cos(theta))/L) b) alpha=3*mg/2*sin(theta)/L ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha Fx=m*ax Fy=m*g*cos(theta)-m*w^2*(L/2) c) cos(theta)=2/3 theta=48.19
i check it r=L/2
bw Fy=m*g*cos(theta)-m*w^2*r what's r???
i know this ,i already done it.thx for the efford!
always here Hill...if you can help with ruler problem it will pe magnificent
yes ROMA V= u*ln(1/(1-p)) a= V/t
sorr that waw 1500*ln(1/0.9750) = 37.91 m/s
no it is wrong!!! he ask you "What is the speed V of the rocket after a burn time of 140 seconds (in m/s)? " that is the half of time and half of % . so .. 1500*ln(1/0.750) = 37.91 m/s a=v/t=37.91/140=0.27
not yet..
excuse me juanpro a=12.83333 f=3.666
juanpro a=12.83333 f=2.333
